Understanding Force Problems: Physics Test Prep and Practice Questions

  • Thread starter Thread starter ruby1
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The discussion focuses on understanding force problems in physics, particularly in preparation for a test. Participants express confusion about applying the correct equations to calculate average force and lift force in two specific scenarios involving a shot-putter and a helicopter. Key equations mentioned include F=ma and the work-energy principle, with suggestions to consider kinetic energy and gravitational force. Clarifications are made regarding the relevance of distance and acceleration in solving these problems. Overall, the conversation emphasizes the need for a solid grasp of fundamental physics concepts to tackle such questions effectively.
ruby1
Messages
1
Reaction score
0
Hi there, I have a physics test tomorrow and I've been trying to review and practice some problems, but I'm not sure I'm understanding it. When the teacher goes through the questions it makes sense, but I'm not always sure where to start. I have a couple questions that I don't understand, even if I could understand one of them it would be good.

Homework Statement


What is the average force exerted by a shot-putter on a 7.0 kg shot if the shot is moved through a distance of 2.8m and is released with a speed of 13 m/s?

So I know
m=7.0 kg
v=13 m/s
d=2.8m
g=9.8 N/kg or 9.8 m/s
and I'm trying to find force.


Homework Equations


a=v/t
F=ma
Not really sure which equations to use.


The Attempt at a Solution



I'm not sure if the distance is even relevant, does it help with the problem at all?
The first thing I thought to do was find the acceleration using a=v/t and then using F=ma, but I don't know the time and wouldn't the acceleration be 0? So if a=o, then F=ma would also be 0, and that doesn't make sense. Overall I'm just not too sure where I would start if I was given a question like this on the test.
The book says the answer is 2.1 x 10N.

Homework Statement


A 6500 kg helicopter accelerates upwards at 0.60 m/s while lifting a 1200 kg car. a) What is the lift force exerted by the air on the rotors? b) What is the tension in the cable (ignore its mass) that connects the car to the helicopter?

total mass=7700kg
a=0.60m

The Attempt at a Solution



So for a) I tried to do F=ma
F= (7700kg) (0.60m) = 4620 N
But the book says the answer is 8.01 x 10^{}4N.
and for b) i honestly have no idea where to start, but the answer is 1.25 x 10^{}4N.

I would really appreciate it if anyone could help me with these problems, thank you very much!
 
Physics news on Phys.org
try using v^2 = v0^2 + 2a(x0 - x) to solve for a
 
First Question:
Work = Force x Distance ( W = Fd )

The amount of work done will result in an equivalent change of Kinetic Energy:

Ek = mv2/2​

Knowing the mass and velocity, you can calculate the kinetic energy, and then force.

ETA:
For your second question, don't forget about the acceleration of gravity (g).
 
Last edited:
Jimmy said:
First Question:
Work = Force x Distance ( W = Fd )

The amount of work done will result in an equivalent change of Kinetic Energy:

Ek = mv2/2​

Knowing the mass and velocity, you can calculate the kinetic energy, and then force.

ETA:
For your second question, don't forget about the acceleration of gravity (g).
do you really need these for this question? I thought that these were just standard f = ma questions
 
are you sure that the answer for B is 833.04 Newtons? it doesn't seem right..
 
wisvuze said:
do you really need these for this question? I thought that these were just standard f = ma questions
No, you're right. I'm just used to doing it that way.

v^2 = 2ax : solve for a

For the second question:
m = 6500kg + 1200kg
a = 9.8m/s^2 + .6 m/s^2

I think the given answer is 8.01x104 N.
But the book says the answer is 8.01 x 10^{}4N.
 
Jimmy said:
No, you're right. I'm just used to doing it that way.

v^2 = 2ax : solve for a

For the second question:
m = 6500kg + 1200kg
a = 9.8m/s^2 + .6 m/s^2

I think the given answer is 8.01x104 N.

Ohh, okay right.. That makes more sense, thanks

I think it should be 0 = Vo^2 + 2a(\Deltax) ? it says that it is 13 m/s when released, right?
 
I notice that you probably want some explanations, since F= (7700kg) (0.60m) = 4620 N as an answer for part B wouldn't really make any sense. You know that after all the forces are calculated, the helicopter moves up with an acceleration of 0.60 m/s^2. So, we know the forces pulling down on the helicopter and the net force - all we need to do now is equate the pulling down forces and some mysterious force (to be known as the rotor air lift force) to be equal to the net force. So if we have forces pulling down on the helicopter, there must exist a force making it go up, since the net force makes it go up
 
Back
Top